{"id":1435,"date":"2018-04-11T22:51:52","date_gmt":"2018-04-12T02:51:52","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/4-4-reaction-yields\/"},"modified":"2018-06-22T23:21:59","modified_gmt":"2018-06-23T03:21:59","slug":"4-4-reaction-yields","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/4-4-reaction-yields\/","title":{"raw":"7.2 Limiting Reagent and Reaction Yields","rendered":"7.2 Limiting Reagent and Reaction Yields"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Explain the concepts of theoretical yield and limiting reactants\/reagents.<\/li>\r\n \t<li>Derive the theoretical yield for a reaction under specified conditions.<\/li>\r\n \t<li>Calculate the percent yield for a reaction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp24998416\">The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as <em>stoichiometric amounts<\/em>. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.<\/p>\r\n\r\n<section id=\"fs-idp5731792\">\r\n<h2>Limiting Reactant<\/h2>\r\n<p id=\"fs-idp103911360\">Consider another food analogy, making grilled cheese sandwiches (<a href=\"#CNX_Chem_04_04_sandwich\" class=\"autogenerated-content\">Figure 1<\/a>):<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp59817360\" style=\"text-align: center\">$latex 1 \\;\\text{slice of cheese} + 2 \\;\\text{slices of bread} \\longrightarrow 1\\;\\text{sandwich}$<\/div>\r\n<p id=\"fs-idp39531056\">Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been <em>limited<\/em> by the number of cheese slices, and the bread slices have been provided in <em>excess<\/em>.<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_04_sandwich\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1300\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_sandwich.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_sandwich-2.jpg\" alt=\"This figure has three rows showing the ingredients needed to make a sandwich. The first row reads, \u201c1 sandwich = 2 slices of bread + 1 slice of cheese.\u201d Two slices of bread and one slice of cheese are shown. The second row reads, \u201cProvided with: 28 slices of bread + 11 slices of cheese.\u201d There are 28 slices of bread and 11 slices of cheese shown. The third row reads, \u201cWe can make: 11 sandwiches + 6 slices of bread left over.\u201d 11 sandwiches are shown with six extra slices of bread.\" width=\"1300\" height=\"719\" \/><\/a> <strong>Figure 1.<\/strong> Sandwich making can illustrate the concepts of limiting and excess reactants.[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<p id=\"fs-idm48112848\">Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp62209424\" style=\"text-align: center\">$latex \\text{H}_2(s) + \\text{Cl}_2(g) \\longrightarrow 2\\text{HCl}(g)$<\/div>\r\n<p id=\"fs-idp157494624\">The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the <strong>limiting reactant<\/strong>, and the other substance is the <strong>excess reactant<\/strong>. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H<sub>2<\/sub> and 2 moles of Cl<sub>2<\/sub>. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.<\/p>\r\n<p id=\"fs-idp22005824\">An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction\u2019s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp67209632\" style=\"text-align: center\">$latex \\text{mol HCl produced} = 3 \\;\\text{mol H}_2 \\times \\frac{2 \\;\\text{mol HCl}}{1 \\;\\text{mol H}_2} = 6 \\;\\text{mol HCl}$<\/div>\r\n<p id=\"fs-idm20022400\">Complete reaction of the provided chlorine would produce<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp10471024\" style=\"text-align: center\">$latex \\text{mol HCl produced} = 2 \\;\\text{mol Cl}_2 \\times \\frac{2 \\;\\text{mol HCl}}{1 \\;\\text{mol Cl}_2} = 4 \\;\\text{mol HCl}$<\/div>\r\n<p id=\"fs-idm39942944\">The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (<a href=\"#CNX_Chem_04_04_limiting\" class=\"autogenerated-content\">Figure 2<\/a>).<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_04_limiting\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1200\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_limiting.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_limiting-2.jpg\" alt=\"The figure shows a space-filling molecular models reacting. There is a reaction arrow pointing to the right in the middle. To the left of the reaction arrow there are three molecules each consisting of two green spheres bonded together. There are also five molecules each consisting of two smaller, white spheres bonded together. Above these molecules is the label, \u201cBefore reaction,\u201d and below these molecules is the label, \u201c6 H subscript 2 and 4 C l subscript 2.\u201d To the right of the reaction arrow, there are eight molecules each consisting of one green sphere bonded to a smaller white sphere. There are also two molecules each consisting of two white spheres bonded together. Above these molecules is the label, \u201cAfter reaction,\u201d and below these molecules is the label, \u201c8 H C l and 2 H subscript 2.\u201d\" width=\"1200\" height=\"545\" \/><\/a> <strong>Figure 2.<\/strong> When H<sub>2<\/sub> and Cl<sub>2<\/sub> are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<div id=\"fs-idp162031984\" class=\"textbox shaded\">\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-5-2.png\" alt=\"\" width=\"151\" height=\"94\" class=\"alignleft\" \/>\r\n\r\n&nbsp;\r\n<p id=\"fs-idm52028528\">View this interactive <a href=\"http:\/\/openstaxcollege.org\/l\/16reactantprod\">simulation<\/a> illustrating the concepts of limiting and excess reactants.<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idp70587344\">\r\n<h3>Example 1<\/h3>\r\n<p id=\"fs-idm3583984\">Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm22587536\" style=\"text-align: center\">$latex 3\\text{Si}(s) + 2\\text{N}_2(g) \\longrightarrow \\text{Si}_3 \\text{N}_4(s)$<\/div>\r\n<p id=\"fs-idp52711328\">Which is the limiting reactant when 2.00 g of Si and 1.50 g of N<sub>2<\/sub> react?<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm18749312\"><strong>Solution<\/strong>\r\nCompute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp16604400\" style=\"text-align: center\">$latex \\text{mol Si} = 2.00 \\;\\rule[0.5ex]{1.75em}{0.1ex}\\hspace{-1.75em}\\text{g Si} \\times \\frac{1 \\;\\text{mol Si}}{28.0855 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{g Si}} = 0.0\\underline{712}11 \\;\\text{mol Si with 3 sig figs}$<\/div>\r\n<div class=\"equation\" id=\"fs-idp42210944\" style=\"text-align: center\">$latex \\text{mol N}_2 = 1.50 \\;\\rule[0.5ex]{2em}{0.1ex}\\hspace{-2em}\\text{g N}_2 \\times \\frac{1 \\;\\text{mol N}_2}{28.0134 \\rule[0.25ex]{1.5em}{0.1ex}\\hspace{-1.5em}\\;\\text{g N}_2} = 0.0\\underline{535}46 \\;\\text{mol N with 3 sig figs}_2$<\/div>\r\n<p id=\"fs-idp47057824\">The provided Si:N<sub>2<\/sub> molar ratio is:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp107313360\" style=\"text-align: center\">$latex \\frac{0.0\\underline{712}11 \\;\\text{mol Si}}{0.0\\underline{535}46 \\;\\text{mol N}_2} = \\frac{1.33 \\;\\text{mol Si}}{1 \\text{mol N}_2}$<\/div>\r\n<p id=\"fs-idm100978944\">The stoichiometric Si:N<sub>2<\/sub> ratio is:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm60202256\" style=\"text-align: center\">$latex \\frac{3 \\;\\text{mol Si}}{2 \\;\\text{mol N}_2} = \\frac{1.5 \\;\\text{mol Si}}{1 \\;\\text{mol N}_2}$<\/div>\r\n<p id=\"fs-idm1915936\">Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.<\/p>\r\n<p id=\"fs-idm9495168\">Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp161973712\">\r\n<p style=\"text-align: center\">$latex \\text{mol Si}_3 \\text{N}_4 \\;\\text{produced} = 0.0\\underline{712}11 \\;\\text{mol Si} \\times \\frac{1 \\;\\text{mol Si}_3 \\text{N}_4}{3 \\;\\text{mol Si}} = 0.0237 \\;\\text{mol Si}_3 \\text{N}_4$<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idm3691696\">while the 0.0535 moles of nitrogen would produce<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm62043536\" style=\"text-align: center\">$latex \\text{mol Si}_3 \\text{N}_4 \\;\\text{produced} = 0.0\\underline{535}46 \\;\\text{mol N}_2 \\times \\frac{1 \\;\\text{mol Si}_3 \\text{N}_4}{2 \\;\\text{mol N}_2} = 0.0268 \\;\\text{mol Si}_3 \\text{N}_4$<\/div>\r\n<p id=\"fs-idm55343248\">Since silicon yields the lesser amount of product, it is the limiting reactant.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp70544208\"><em><strong>Test Yourself<\/strong><\/em>\r\nWhich is the limiting reactant when 5.00 g of H<sub>2<\/sub> and 10.0 g of O<sub>2<\/sub> react and form water?<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\nO<sub>2<\/sub>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm20935792\">\r\n<h2>Percent Yield<\/h2>\r\n<p id=\"fs-idm22072192\">The amount of product that <em>may be<\/em> produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the <strong>theoretical yield<\/strong> of the reaction. In practice, the amount of product obtained is called the <strong>actual yield<\/strong>, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by <em>side reactions<\/em> that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction\u2019s theoretical yield is achieved is commonly expressed as its <strong>percent yield<\/strong>:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp47653952\" style=\"text-align: center\">$latex \\text{percent yield} = \\frac{\\text{actual yield}}{\\text{theoretical yield}} \\times 100\\% $<\/div>\r\n<p id=\"fs-idm52282816\">Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.<\/p>\r\n\r\n<div class=\"textbox shaded\" id=\"fs-idm49018784\">\r\n<h3>Example 2<\/h3>\r\n<p id=\"fs-idm68646768\">Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp101757552\" style=\"text-align: center\">$latex \\text{CuSO}_4(aq) + \\text{Zn}(s) \\longrightarrow \\text{Cu}(s) + \\text{ZnSO}_4(aq)$<\/div>\r\n<p id=\"fs-idp104261712\">What is the percent yield?<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp98890704\"><strong>Solution<\/strong>\r\nThe provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp104521856\">\r\n<p style=\"text-align: center\">$latex 1.274 \\;\\rule[0.5ex]{3.75em}{0.1ex}\\hspace{-3.75em}\\text{g CuSO}_4 \\times \\frac{1 \\;\\rule[0.25ex]{4em}{0.1ex}\\hspace{-4em}\\text{mol CuSO}_4}{159.610 \\;\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em}\\text{g CuSO}_4} \\times \\frac{1 \\;\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol Cu}}{1 \\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol CuSO}_4} \\times \\frac{63.546 \\;\\text{g Cu}}{1 \\;\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol Cu}} = 0.\\underline{5072}21 \\;\\text{g Cu with 4 sig figs}$<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idp47540336\">Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp217309712\" style=\"text-align: center\">$latex \\text{percent yield} = (\\frac{\\text{actual yield}}{\\text{theoretical yield}}) \\times 100\\%$<\/div>\r\n<div class=\"equation\" id=\"fs-idp182503696\">\r\n<p style=\"text-align: center\">$latex \\text{percent yield} = (\\frac{0.392 \\;\\text{g Cu}}{0.\\underline{5072}21 \\;\\text{g Cu}}) \\times 100\\%$<\/p>\r\n<p style=\"text-align: center\">$latex = 77.3\\%$<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idp38988928\"><em><strong>Test Yourself<\/strong><\/em>\r\nWhat is the percent yield of a reaction that produces 12.5 g of the gas Freon CF<sub>2<\/sub>Cl<sub>2<\/sub> from 32.9 g of CCl<sub>4<\/sub> and excess HF?<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp61435552\" style=\"text-align: center\">$latex \\text{CCl}_4 + 2\\text{HF} \\longrightarrow \\text{CF}_2 \\text{Cl}_2 + 2\\text{HCl}$<\/div>\r\n<div><\/div>\r\n<div><em><strong>Answer<\/strong><\/em><\/div>\r\n<div>48.3%<\/div>\r\n<\/div>\r\n<div id=\"fs-idm21958480\" class=\"textbox shaded\">\r\n<h3 class=\"title\">Green Chemistry and Atom Economy<\/h3>\r\n<p id=\"fs-idm49080288\">The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as <em>green chemistry<\/em>. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the \u201cTwelve Principles of Green Chemistry\u201d (see details at this <a href=\"http:\/\/openstaxcollege.org\/l\/16greenchem\">website<\/a>). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The <em>atom economy<\/em> of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of <em>all<\/em> the reactants used:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp116950848\" style=\"text-align: center\">$latex \\text{atom economy} = \\frac{\\text{mass of product}}{\\text{mass of reactants}} \\times 100\\% $<\/div>\r\n<p id=\"fs-idp59920960\">Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the <em>theoretical<\/em> efficiencies of <em>different<\/em> chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.<\/p>\r\n<p id=\"fs-idp167996000\">The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (<a href=\"#CNX_Chem_04_04_GreenChem\" class=\"autogenerated-content\">Figure 3<\/a>). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%.<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_04_GreenChem\"><figcaption>[caption id=\"\" align=\"aligncenter\" width=\"427\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_GreenChem.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_GreenChem-2.jpg\" alt=\"This figure is labeled, \u201ca,\u201d and, \u201cb.\u201d Part a shows an open bottle of ibuprofen and a small pile of ibuprofen tablets beside it. Part b shows a reaction along with line structures. The first line structure looks like a diagonal line pointing down and to the right, then up and to the right and then down and to the right. At this point it connects to a hexagon with alternating double bonds. At the first trough there is a line that points straight down. From this structure, there is an arrow pointing downward. The arrow is labeled, \u201cH F,\u201d on the left and \u201c( C H subscript 3 C O ) subscript 2 O,\u201d on the right. The next line structure looks exactly like the first line structure, but it has a line angled down and to the right from the lower right point of the hexagon. This line is connected to another line which points straight down. Where these two lines meet, there is a double bond to an O atom. There is another arrow pointing downward, and it is labeled, \u201cH subscript 2, Raney N i.\u201d The next structure looks very similar to the second, previous structure, except in place of the double bonded O, there is a singly bonded O H group. There is a final reaction arrow pointing downward, and it is labeled, \u201cC O, [ P d ].\u201d The final structure is similar to the third, previous structure except in place of the O H group, there is another line that points down and to the right to an O H group. At these two lines, there is a double bonded O.\" width=\"427\" height=\"378\" class=\"\" \/><\/a> <strong>Figure 3.<\/strong> (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)[\/caption]In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency\u2019s Greener Synthetic Pathways Award in 1997.<\/figcaption><\/figure>\r\n<\/div>\r\n<\/section><section id=\"fs-idm64875168\" class=\"summary\">\r\n<h2>More Worked Out Problems<\/h2>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 3<\/h3>\r\n<p id=\"ball-ch05_s06_p13\" class=\"para\">A 5.00 g quantity of Rb are combined with 3.44 g of MgCl<sub class=\"subscript\">2<\/sub> according to this chemical reaction:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 Rb(s) +\u00a0MgCl<sub class=\"subscript\">2<\/sub>(s) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Mg(s) +\u00a02 RbCl(s)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s06_p14\" class=\"para\">What mass of Mg is formed, and what mass of what reactant is left over?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s06_p15\" class=\"para\">Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less.<\/p>\r\n<p style=\"text-align: center\">$latex 5.00 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}}{85.4678 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb}} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}}{2 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}} \\times \\frac{24.3050 \\;\\text{g Mg}}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}} = 0.\\underline{711}53 \\;\\text{g Mg with 3 sig figs}$<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center\">$latex 3.44 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g MgCl}_2 \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2}{95.2104 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g MgCl}_2} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}}{1 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2} \\times \\frac{24.3050 \\;\\text{g Mg}}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}} = 0.\\underline{878}15 \\;\\text{g Mg with 3 sig figs}$<\/p>\r\n&nbsp;\r\n<p id=\"ball-ch05_s06_p16\" class=\"para\">The 0.712 g of Mg is the lesser quantity, so the associated reactant\u20145.00 g of Rb\u2014is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl<sub class=\"subscript\">2<\/sub> reacted with the 5.00 g of Rb and then subtract the amount reacted from the original amount.<\/p>\r\n$latex 5.00 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}}{85.4678 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb}} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2}{2 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}} \\times \\frac{95.2104 \\;\\text{g MgCl}_2}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2} = \\underline{2.78}498 \\;\\text{g magnesium chloride with 3 sig figs}$\r\n\r\n&nbsp;\r\n<p id=\"ball-ch05_s06_p17\" class=\"para\">Because we started with 3.44 g of MgCl<sub class=\"subscript\">2<\/sub>, we have<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">3.44 g MgCl<sub class=\"subscript\">2<\/sub> \u2212 \u00a0<span style=\"text-decoration: underline\">2.78<\/span>498 g MgCl<sub class=\"subscript\">2<\/sub> reacted = 0.66 g MgCl<sub class=\"subscript\">2<\/sub> left<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s06_p18\" class=\"para\">Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent?<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/MGOS.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/MGOS-1.png\" alt=\"MGOS\" width=\"351\" height=\"68\" class=\"wp-image-3752 aligncenter\" \/><\/a>\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s06_p19\" class=\"para\">H<sub class=\"subscript\">2<\/sub>S is the limiting reagent; 1.5 g of MgO are left over.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 4<\/h3>\r\n<p id=\"ball-ch05_s05_p06\" class=\"para\">A worker reacts 30.5 g of Zn with nitric acid and evapourates the remaining water to obtain 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. What are the theoretical yield, the actual yield, and the percent yield?<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 HNO<sub class=\"subscript\">3<\/sub>(aq) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s05_p07\" class=\"para\">A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.37 g\/mol) and Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> (189.38 g\/mol). In three steps, the mass-mass calculation is<\/p>\r\n$latex 30.5 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Zn} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn}}{65.37 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Zn}} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn(NO}_3 \\text{)}_2}{1 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn}} \\times \\frac{189.38 \\;\\text{g Zn(NO}_3 \\text{)}_2}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn(NO}_3 \\text{)}_2} = \\underline{88.3}6 \\;\\text{g zing nitrate with 3 sig figs}$\r\n<p id=\"ball-ch05_s05_p08\" class=\"para\">Thus, the theoretical yield is 88.4 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:<\/p>\r\n<p style=\"text-align: center\">$latex \\text{percent yield} = (\\frac{65.2 \\;\\text{g Zn(NO}_3 \\text{)}_2}{\\underline{88.3}6 \\;\\text{g Zn(NO}_3 \\text{)}_2}) \\times 100\\%$<\/p>\r\n<p style=\"text-align: center\">$latex = 73.8\\%$<\/p>\r\n<p id=\"ball-ch05_s05_p09\" class=\"para\">The worker achieved almost three-fourths of the possible yield.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s05_p10\" class=\"para\">A synthesis produced 2.05 g of NH<sub class=\"subscript\">3<\/sub> from 16.5 g of N<sub class=\"subscript\">2<\/sub>. What is the theoretical yield and the percent yield?<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s05_p11\" class=\"para\">theoretical yield = 20.1 g; percent yield = 10.2%<\/p>\r\n\r\n<\/div>\r\n<div class=\"callout block\" id=\"ball-ch05_s05_n03\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification<\/h3>\r\n<p id=\"ball-ch05_s05_p12\" class=\"para\">Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.<\/p>\r\n<p id=\"ball-ch05_s05_p13\" class=\"para\">Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C<sub class=\"subscript\">13<\/sub>H<sub class=\"subscript\">21<\/sub>NO<sub class=\"subscript\">2<\/sub>; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:<\/p>\r\n\r\n<div class=\"informaltable\">\r\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td>impure albuterol \u2192\u00a0intermediate A<\/td>\r\n<td>percent yield = 70%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate A \u2192\u00a0intermediate B<\/td>\r\n<td>percent yield = 100%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate B \u2192\u00a0intermediate C<\/td>\r\n<td>percent yield = 40%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate C \u2192\u00a0intermediate D<\/td>\r\n<td>percent yield = 72%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate D \u2192\u00a0purified albuterol<\/td>\r\n<td>percent yield = 35%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">overall percent yield = 70% \u00d7 100% \u00d7 40% \u00d7 72% \u00d7 35% = 7.5%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<p id=\"ball-ch05_s05_p14\" class=\"para\">That is, only about <em class=\"emphasis\">one-fourteenth<\/em> of the original material was turned into the purified drug. This gives you one reason why some drugs are so expensive; a lot of material is lost in making a high-purity pharmaceutical.<\/p>\r\n\r\n<\/div>\r\n<h2 class=\"para\">Key Concepts and Summary<\/h2>\r\n<\/div>\r\n<p id=\"fs-idm5314032\">When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.<\/p>\r\n\r\n<\/section><section id=\"fs-idp36036608\" class=\"key-equations\">\r\n<h2>Key Equations<\/h2>\r\n<ul id=\"fs-idp46607520\">\r\n \t<li>$latex \\text{percent yield} = (\\frac{\\text{actual yield}}{\\text{theoretical yield}}) \\times 100\\%$<\/li>\r\n<\/ul>\r\n<\/section><section id=\"fs-idm49138160\" class=\"exercises\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n1. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?\r\n\r\n2. A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was his percent yield?\r\n\r\n3. Freon-12, CCl<sub>2<\/sub>F<sub>2<\/sub>, is prepared from CCl<sub>4<\/sub> by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl<sub>2<\/sub>F<sub>2<\/sub> from 32.9 g of CCl<sub>4<\/sub>. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield.\r\n\r\n4. Toluene, C<sub>6<\/sub>H<sub>5<\/sub>CH<sub>3<\/sub>, is oxidized by air under carefully controlled conditions to benzoic acid, C<sub>6<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>H, which is used to prepare the food preservative sodium benzoate, C<sub>6<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?\r\n<p id=\"fs-idp58383920\" style=\"text-align: center\">$latex 2\\text{C}_6\\text{H}_5\\text{CH}_3\\;+\\;3\\text{O}_2\\;{\\longrightarrow}\\;2\\text{C}_6\\text{H}_5\\text{CO}_2\\text{H}\\;+\\;2\\text{H}_2\\text{O}$<\/p>\r\n5. Outline the steps needed to solve the following problem, then do the calculations. Ether, (C<sub>2<\/sub>H<sub>5<\/sub>)<sub>2<\/sub>O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.\r\n<p id=\"fs-idm120301824\" style=\"text-align: center\">$latex 2\\text{C}_2\\text{H}_5\\text{OH}\\;+\\;\\text{H}_2\\text{SO}_4\\;{\\longrightarrow}\\;(\\text{C}_2\\text{H}_5)_2\\;+\\;\\text{H}_2\\text{SO}_4{\\cdot}\\text{H}_2\\text{O}$<\/p>\r\n<p id=\"fs-idp39307056\">What is the percent yield of ether if 1.17 L (d = 0.7134 g\/mL) is isolated from the reaction of 1.500 L of C<sub>2<\/sub>H<sub>5<\/sub>OH\u00a0(d = 0.7894 g\/mL)?<\/p>\r\n6. Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H<sub>3<\/sub>PO<sub>4<\/sub> react according to the following chemical equation.\r\n<p style=\"text-align: center\">$latex 2\\text{Cr}\\;+\\;2\\text{H}_3\\text{PO}_4\\;{\\longrightarrow}\\;2\\text{CrPO}_4\\;+\\;3\\text{H}_2$<\/p>\r\n<p id=\"fs-idm522144\">Determine the limiting reactant.<\/p>\r\n7. Uranium can be isolated from its ores by dissolving it as UO<sub>2<\/sub>(NO<sub>3<\/sub>)<sub>2<\/sub>, then separating it as solid UO<sub>2<\/sub>(C<sub>2<\/sub>O<sub>4<\/sub>)\u00b73H<sub>2<\/sub>O. Addition of 0.4031 g of sodium oxalate, Na<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub>, to a solution containing 1.481 g of uranyl nitrate, UO<sub>2<\/sub>(NO<sub>3<\/sub>)<sub>2<\/sub>, yields 1.073 g of solid UO<sub>2<\/sub>(C<sub>2<\/sub>O<sub>4<\/sub>)\u00b73H<sub>2<\/sub>O.\r\n<p id=\"fs-idm72766640\" style=\"text-align: center\">$latex \\text{Na}_2\\text{C}_2\\text{O}_4\\;+\\;\\text{UO}_2(\\text{NO}_3)_2\\;+\\;3\\text{H}_2\\text{O}\\;{\\longrightarrow}\\;\\text{UO}_2(\\text{C}_2\\text{O}_4){\\cdot}3\\text{H}_2\\text{O}\\;+\\;2\\text{NaNO}_3$<\/p>\r\n<p id=\"fs-idp232896464\">Determine the limiting reactant and the percent yield of this reaction.<\/p>\r\n8. How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_saccharin_img-2.jpg\" alt=\"A structural formula is shown. A hexagonal ring of 6 C atoms with alternating double bonds has single H atoms bonded to four consecutive C atoms on the left side of the ring. The two C atoms on the right side of the ring, which are joined by a double bond, are also included in a 5 atom ring to their right. The C atom of this pair that is nearest the top of the structure is singly bonded to a C atom at the top of the 5 atom ring which has an O atom double bonded above. An N atom is singly bonded to the lower right of this same C atom. The N atom has an H atom bonded to its right and to its lower left, it is bonded to an S atom. The S atom is connected to the second C atom that is shared in the two rings. The S atom is also double bonded to an O atom to its lower right and is double bonded to a second O atom directly below it.\" width=\"198\" height=\"155\" class=\"aligncenter\" \/>\r\n\r\n9. Would you agree to buy 1 trillion (1,000,000,000,000) gold atoms for $5? Explain why or why not. Find the current price of gold at http:\/\/money.cnn.com\/data\/commodities\/ (1 troy ounce = 31.1 g)\r\n<div class=\"qandaset block\" id=\"ball-ch05_s06_qs01\">\r\n\r\n10. The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH<sub class=\"subscript\">3<\/sub>. What is the limiting reagent?\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Limiting-Reagent.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Limiting-Reagent-1.png\" alt=\"Limiting Reagent\" width=\"276\" height=\"284\" class=\"wp-image-4659 aligncenter\" \/><\/a>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<div class=\"informalfigure medium\"><span style=\"font-size: 1em\">11. Given the statement \u201c20.0 g of methane is burned in excess oxygen,\u201d is it obvious which reactant is the limiting reagent?<\/span><\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">12. Acetylene (C<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">) is formed by reacting 7.08 g of C and 4.92 g of H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">.<\/span><\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 C(s) +\u00a0H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s06_qs01_p10\" class=\"para\">What is the limiting reagent? How much of the other reactant is in excess?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">13. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?<\/span><\/div>\r\n<div class=\"question\">\r\n\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-22-at-1.49.34-PM.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-22-at-1.49.34-PM-1.png\" alt=\"Screen Shot 2014-07-22 at 1.49.34 PM\" width=\"244\" height=\"49\" class=\"wp-image-3754 aligncenter\" \/><\/a>\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">14. To form the precipitate PbCl<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">, 2.88 g of NaCl and 7.21 g of Pb(NO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">)<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?<\/span><\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">15. What is the difference between the theoretical yield and the actual yield?<\/span><\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">16. A worker isolates 2.675 g of SiF<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\"> after reacting 2.339 g of SiO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> with HF. What are the theoretical yield and the actual yield?<\/span><\/div>\r\n<div class=\"question\">\r\n<div class=\"qandaset block\" id=\"ball-ch05_s05_qs01\">\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">SiO<sub class=\"subscript\">2<\/sub>(s) +\u00a04 HF(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0SiF<sub class=\"subscript\">4<\/sub>(g) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">17. A chemist decomposes 1.006 g of NaHCO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> and obtains 0.0334 g of Na<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">CO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">. What are the theoretical yield and the actual yield?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 NaHCO<sub class=\"subscript\">3<\/sub>(s) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0CO<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">18. What is the percent yield in Exercise 16?<\/span>\r\n\r\n<span style=\"font-size: 1em\">19. What is the percent yield in Exercise 17?<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<strong>Answers<\/strong>\r\n<p id=\"fs-idp22297728\">1. The limiting reactant is Cl<sub>2<\/sub>.<\/p>\r\n<p id=\"fs-idp46323152\">2. Percent yield = 31%<\/p>\r\n<p id=\"fs-idp59773152\">3. g CCl<sub>4<\/sub>\u00a0<span class=\"informalequation\"><span class=\"mathphrase\"><span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span><\/span><\/span> mol CCl<sub>4<\/sub>\u00a0<span class=\"informalequation\"><span class=\"mathphrase\"><span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span><\/span><\/span> mol CCl<sub>2<\/sub>F<sub>2<\/sub>\u00a0<span class=\"informalequation\"><span class=\"mathphrase\"><span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span><\/span><\/span> g CCl<sub>2<\/sub>F<sub>2<\/sub>, percent yield = 48.3%<\/p>\r\n<p id=\"fs-idm18984464\">4. percent yield = 91.3%<\/p>\r\n<p id=\"fs-idm48114608\">5. Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6%<\/p>\r\n<p id=\"fs-idm38791728\">6. The conversion needed is $latex \\text{mol\\;Cr}\\;{\\longrightarrow}\\;\\text{mol\\;H}_3\\text{PO}_4$. Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.<\/p>\r\n<p id=\"fs-idp74993536\">7. Na<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub> is the limiting reactant. percent yield = 86.6%<\/p>\r\n<p id=\"fs-idp81228400\">8. Only four molecules can be made.<\/p>\r\n<p id=\"fs-idp8960064\">9. This amount cannot be weighted by ordinary balances and is worthless.<\/p>\r\n10. Nitrogen is the limiting reagent.\r\n\r\n11.\u00a0Yes; methane is the limiting reagent.\r\n\r\n12.\u00a0C is the limiting reagent; 4.33 g of H<sub class=\"subscript\">2<\/sub> are left over.\r\n\r\n13.\u00a0H<sub class=\"subscript\">2<\/sub>O is the limiting reagent; 25.9 g of P<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">6<\/sub> are left over.\r\n\r\n14.\u00a06.06 g of PbCl<sub class=\"subscript\">2<\/sub> are formed; 0.33 g of NaCl is left over.\r\n\r\n15. Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.\r\n\r\n16.\u00a0theoretical yield = 4.052 g; actual yield = 2.675 g\r\n\r\n17.\u00a0theoretical yield = 0.635 g; actual yield = 0.0334 g\r\n\r\n18.\u00a066.02%\r\n\r\n19.\u00a05.26%\r\n\r\n<\/div>\r\n<\/section>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<strong>actual yield:\u00a0<\/strong>amount of product formed in a reaction\r\n\r\n<strong>excess reactant:\u00a0<\/strong>reactant present in an amount greater than required by the reaction stoichiometry\r\n\r\n<strong>limiting reactant:\u00a0<\/strong>reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated\r\n\r\n<strong>percent yield:\u00a0<\/strong>measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield\r\n\r\n<strong>theoretical yield:\u00a0<\/strong>amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Explain the concepts of theoretical yield and limiting reactants\/reagents.<\/li>\n<li>Derive the theoretical yield for a reaction under specified conditions.<\/li>\n<li>Calculate the percent yield for a reaction.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp24998416\">The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as <em>stoichiometric amounts<\/em>. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.<\/p>\n<section id=\"fs-idp5731792\">\n<h2>Limiting Reactant<\/h2>\n<p id=\"fs-idp103911360\">Consider another food analogy, making grilled cheese sandwiches (<a href=\"#CNX_Chem_04_04_sandwich\" class=\"autogenerated-content\">Figure 1<\/a>):<\/p>\n<div class=\"equation\" id=\"fs-idp59817360\" style=\"text-align: center\">[latex]1 \\;\\text{slice of cheese} + 2 \\;\\text{slices of bread} \\longrightarrow 1\\;\\text{sandwich}[\/latex]<\/div>\n<p id=\"fs-idp39531056\">Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been <em>limited<\/em> by the number of cheese slices, and the bread slices have been provided in <em>excess<\/em>.<\/p>\n<figure id=\"CNX_Chem_04_04_sandwich\"><figcaption>\n<figure style=\"width: 1300px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_sandwich.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_sandwich-2.jpg\" alt=\"This figure has three rows showing the ingredients needed to make a sandwich. The first row reads, \u201c1 sandwich = 2 slices of bread + 1 slice of cheese.\u201d Two slices of bread and one slice of cheese are shown. The second row reads, \u201cProvided with: 28 slices of bread + 11 slices of cheese.\u201d There are 28 slices of bread and 11 slices of cheese shown. The third row reads, \u201cWe can make: 11 sandwiches + 6 slices of bread left over.\u201d 11 sandwiches are shown with six extra slices of bread.\" width=\"1300\" height=\"719\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> Sandwich making can illustrate the concepts of limiting and excess reactants.<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<p id=\"fs-idm48112848\">Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:<\/p>\n<div class=\"equation\" id=\"fs-idp62209424\" style=\"text-align: center\">[latex]\\text{H}_2(s) + \\text{Cl}_2(g) \\longrightarrow 2\\text{HCl}(g)[\/latex]<\/div>\n<p id=\"fs-idp157494624\">The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the <strong>limiting reactant<\/strong>, and the other substance is the <strong>excess reactant<\/strong>. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H<sub>2<\/sub> and 2 moles of Cl<sub>2<\/sub>. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.<\/p>\n<p id=\"fs-idp22005824\">An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction\u2019s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield<\/p>\n<div class=\"equation\" id=\"fs-idp67209632\" style=\"text-align: center\">[latex]\\text{mol HCl produced} = 3 \\;\\text{mol H}_2 \\times \\frac{2 \\;\\text{mol HCl}}{1 \\;\\text{mol H}_2} = 6 \\;\\text{mol HCl}[\/latex]<\/div>\n<p id=\"fs-idm20022400\">Complete reaction of the provided chlorine would produce<\/p>\n<div class=\"equation\" id=\"fs-idp10471024\" style=\"text-align: center\">[latex]\\text{mol HCl produced} = 2 \\;\\text{mol Cl}_2 \\times \\frac{2 \\;\\text{mol HCl}}{1 \\;\\text{mol Cl}_2} = 4 \\;\\text{mol HCl}[\/latex]<\/div>\n<p id=\"fs-idm39942944\">The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (<a href=\"#CNX_Chem_04_04_limiting\" class=\"autogenerated-content\">Figure 2<\/a>).<\/p>\n<figure id=\"CNX_Chem_04_04_limiting\"><figcaption>\n<figure style=\"width: 1200px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_limiting.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_limiting-2.jpg\" alt=\"The figure shows a space-filling molecular models reacting. There is a reaction arrow pointing to the right in the middle. To the left of the reaction arrow there are three molecules each consisting of two green spheres bonded together. There are also five molecules each consisting of two smaller, white spheres bonded together. Above these molecules is the label, \u201cBefore reaction,\u201d and below these molecules is the label, \u201c6 H subscript 2 and 4 C l subscript 2.\u201d To the right of the reaction arrow, there are eight molecules each consisting of one green sphere bonded to a smaller white sphere. There are also two molecules each consisting of two white spheres bonded together. Above these molecules is the label, \u201cAfter reaction,\u201d and below these molecules is the label, \u201c8 H C l and 2 H subscript 2.\u201d\" width=\"1200\" height=\"545\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong> When H<sub>2<\/sub> and Cl<sub>2<\/sub> are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<div id=\"fs-idp162031984\" class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-5-2.png\" alt=\"\" width=\"151\" height=\"94\" class=\"alignleft\" \/><\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm52028528\">View this interactive <a href=\"http:\/\/openstaxcollege.org\/l\/16reactantprod\">simulation<\/a> illustrating the concepts of limiting and excess reactants.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox shaded\" id=\"fs-idp70587344\">\n<h3>Example 1<\/h3>\n<p id=\"fs-idm3583984\">Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:<\/p>\n<div class=\"equation\" id=\"fs-idm22587536\" style=\"text-align: center\">[latex]3\\text{Si}(s) + 2\\text{N}_2(g) \\longrightarrow \\text{Si}_3 \\text{N}_4(s)[\/latex]<\/div>\n<p id=\"fs-idp52711328\">Which is the limiting reactant when 2.00 g of Si and 1.50 g of N<sub>2<\/sub> react?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm18749312\"><strong>Solution<\/strong><br \/>\nCompute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.<\/p>\n<div class=\"equation\" id=\"fs-idp16604400\" style=\"text-align: center\">[latex]\\text{mol Si} = 2.00 \\;\\rule[0.5ex]{1.75em}{0.1ex}\\hspace{-1.75em}\\text{g Si} \\times \\frac{1 \\;\\text{mol Si}}{28.0855 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{g Si}} = 0.0\\underline{712}11 \\;\\text{mol Si with 3 sig figs}[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idp42210944\" style=\"text-align: center\">[latex]\\text{mol N}_2 = 1.50 \\;\\rule[0.5ex]{2em}{0.1ex}\\hspace{-2em}\\text{g N}_2 \\times \\frac{1 \\;\\text{mol N}_2}{28.0134 \\rule[0.25ex]{1.5em}{0.1ex}\\hspace{-1.5em}\\;\\text{g N}_2} = 0.0\\underline{535}46 \\;\\text{mol N with 3 sig figs}_2[\/latex]<\/div>\n<p id=\"fs-idp47057824\">The provided Si:N<sub>2<\/sub> molar ratio is:<\/p>\n<div class=\"equation\" id=\"fs-idp107313360\" style=\"text-align: center\">[latex]\\frac{0.0\\underline{712}11 \\;\\text{mol Si}}{0.0\\underline{535}46 \\;\\text{mol N}_2} = \\frac{1.33 \\;\\text{mol Si}}{1 \\text{mol N}_2}[\/latex]<\/div>\n<p id=\"fs-idm100978944\">The stoichiometric Si:N<sub>2<\/sub> ratio is:<\/p>\n<div class=\"equation\" id=\"fs-idm60202256\" style=\"text-align: center\">[latex]\\frac{3 \\;\\text{mol Si}}{2 \\;\\text{mol N}_2} = \\frac{1.5 \\;\\text{mol Si}}{1 \\;\\text{mol N}_2}[\/latex]<\/div>\n<p id=\"fs-idm1915936\">Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.<\/p>\n<p id=\"fs-idm9495168\">Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield<\/p>\n<div class=\"equation\" id=\"fs-idp161973712\">\n<p style=\"text-align: center\">[latex]\\text{mol Si}_3 \\text{N}_4 \\;\\text{produced} = 0.0\\underline{712}11 \\;\\text{mol Si} \\times \\frac{1 \\;\\text{mol Si}_3 \\text{N}_4}{3 \\;\\text{mol Si}} = 0.0237 \\;\\text{mol Si}_3 \\text{N}_4[\/latex]<\/p>\n<\/div>\n<p id=\"fs-idm3691696\">while the 0.0535 moles of nitrogen would produce<\/p>\n<div class=\"equation\" id=\"fs-idm62043536\" style=\"text-align: center\">[latex]\\text{mol Si}_3 \\text{N}_4 \\;\\text{produced} = 0.0\\underline{535}46 \\;\\text{mol N}_2 \\times \\frac{1 \\;\\text{mol Si}_3 \\text{N}_4}{2 \\;\\text{mol N}_2} = 0.0268 \\;\\text{mol Si}_3 \\text{N}_4[\/latex]<\/div>\n<p id=\"fs-idm55343248\">Since silicon yields the lesser amount of product, it is the limiting reactant.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp70544208\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWhich is the limiting reactant when 5.00 g of H<sub>2<\/sub> and 10.0 g of O<sub>2<\/sub> react and form water?<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>O<sub>2<\/sub><\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm20935792\">\n<h2>Percent Yield<\/h2>\n<p id=\"fs-idm22072192\">The amount of product that <em>may be<\/em> produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the <strong>theoretical yield<\/strong> of the reaction. In practice, the amount of product obtained is called the <strong>actual yield<\/strong>, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by <em>side reactions<\/em> that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction\u2019s theoretical yield is achieved is commonly expressed as its <strong>percent yield<\/strong>:<\/p>\n<div class=\"equation\" id=\"fs-idp47653952\" style=\"text-align: center\">[latex]\\text{percent yield} = \\frac{\\text{actual yield}}{\\text{theoretical yield}} \\times 100\\%[\/latex]<\/div>\n<p id=\"fs-idm52282816\">Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.<\/p>\n<div class=\"textbox shaded\" id=\"fs-idm49018784\">\n<h3>Example 2<\/h3>\n<p id=\"fs-idm68646768\">Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:<\/p>\n<div class=\"equation\" id=\"fs-idp101757552\" style=\"text-align: center\">[latex]\\text{CuSO}_4(aq) + \\text{Zn}(s) \\longrightarrow \\text{Cu}(s) + \\text{ZnSO}_4(aq)[\/latex]<\/div>\n<p id=\"fs-idp104261712\">What is the percent yield?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp98890704\"><strong>Solution<\/strong><br \/>\nThe provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:<\/p>\n<div class=\"equation\" id=\"fs-idp104521856\">\n<p style=\"text-align: center\">[latex]1.274 \\;\\rule[0.5ex]{3.75em}{0.1ex}\\hspace{-3.75em}\\text{g CuSO}_4 \\times \\frac{1 \\;\\rule[0.25ex]{4em}{0.1ex}\\hspace{-4em}\\text{mol CuSO}_4}{159.610 \\;\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em}\\text{g CuSO}_4} \\times \\frac{1 \\;\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol Cu}}{1 \\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol CuSO}_4} \\times \\frac{63.546 \\;\\text{g Cu}}{1 \\;\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol Cu}} = 0.\\underline{5072}21 \\;\\text{g Cu with 4 sig figs}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-idp47540336\">Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be<\/p>\n<div class=\"equation\" id=\"fs-idp217309712\" style=\"text-align: center\">[latex]\\text{percent yield} = (\\frac{\\text{actual yield}}{\\text{theoretical yield}}) \\times 100\\%[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idp182503696\">\n<p style=\"text-align: center\">[latex]\\text{percent yield} = (\\frac{0.392 \\;\\text{g Cu}}{0.\\underline{5072}21 \\;\\text{g Cu}}) \\times 100\\%[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]= 77.3\\%[\/latex]<\/p>\n<\/div>\n<p id=\"fs-idp38988928\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWhat is the percent yield of a reaction that produces 12.5 g of the gas Freon CF<sub>2<\/sub>Cl<sub>2<\/sub> from 32.9 g of CCl<sub>4<\/sub> and excess HF?<\/p>\n<div class=\"equation\" id=\"fs-idp61435552\" style=\"text-align: center\">[latex]\\text{CCl}_4 + 2\\text{HF} \\longrightarrow \\text{CF}_2 \\text{Cl}_2 + 2\\text{HCl}[\/latex]<\/div>\n<div><\/div>\n<div><em><strong>Answer<\/strong><\/em><\/div>\n<div>48.3%<\/div>\n<\/div>\n<div id=\"fs-idm21958480\" class=\"textbox shaded\">\n<h3 class=\"title\">Green Chemistry and Atom Economy<\/h3>\n<p id=\"fs-idm49080288\">The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as <em>green chemistry<\/em>. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the \u201cTwelve Principles of Green Chemistry\u201d (see details at this <a href=\"http:\/\/openstaxcollege.org\/l\/16greenchem\">website<\/a>). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The <em>atom economy<\/em> of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of <em>all<\/em> the reactants used:<\/p>\n<div class=\"equation\" id=\"fs-idp116950848\" style=\"text-align: center\">[latex]\\text{atom economy} = \\frac{\\text{mass of product}}{\\text{mass of reactants}} \\times 100\\%[\/latex]<\/div>\n<p id=\"fs-idp59920960\">Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the <em>theoretical<\/em> efficiencies of <em>different<\/em> chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.<\/p>\n<p id=\"fs-idp167996000\">The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (<a href=\"#CNX_Chem_04_04_GreenChem\" class=\"autogenerated-content\">Figure 3<\/a>). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%.<\/p>\n<figure id=\"CNX_Chem_04_04_GreenChem\"><figcaption>\n<figure style=\"width: 427px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_GreenChem.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_GreenChem-2.jpg\" alt=\"This figure is labeled, \u201ca,\u201d and, \u201cb.\u201d Part a shows an open bottle of ibuprofen and a small pile of ibuprofen tablets beside it. Part b shows a reaction along with line structures. The first line structure looks like a diagonal line pointing down and to the right, then up and to the right and then down and to the right. At this point it connects to a hexagon with alternating double bonds. At the first trough there is a line that points straight down. From this structure, there is an arrow pointing downward. The arrow is labeled, \u201cH F,\u201d on the left and \u201c( C H subscript 3 C O ) subscript 2 O,\u201d on the right. The next line structure looks exactly like the first line structure, but it has a line angled down and to the right from the lower right point of the hexagon. This line is connected to another line which points straight down. Where these two lines meet, there is a double bond to an O atom. There is another arrow pointing downward, and it is labeled, \u201cH subscript 2, Raney N i.\u201d The next structure looks very similar to the second, previous structure, except in place of the double bonded O, there is a singly bonded O H group. There is a final reaction arrow pointing downward, and it is labeled, \u201cC O, [ P d ].\u201d The final structure is similar to the third, previous structure except in place of the O H group, there is another line that points down and to the right to an O H group. At these two lines, there is a double bonded O.\" width=\"427\" height=\"378\" class=\"\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)<\/figcaption><\/figure>\n<p>In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency\u2019s Greener Synthetic Pathways Award in 1997.<\/figcaption><\/figure>\n<\/div>\n<\/section>\n<section id=\"fs-idm64875168\" class=\"summary\">\n<h2>More Worked Out Problems<\/h2>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch05_s06_p13\" class=\"para\">A 5.00 g quantity of Rb are combined with 3.44 g of MgCl<sub class=\"subscript\">2<\/sub> according to this chemical reaction:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 Rb(s) +\u00a0MgCl<sub class=\"subscript\">2<\/sub>(s) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Mg(s) +\u00a02 RbCl(s)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s06_p14\" class=\"para\">What mass of Mg is formed, and what mass of what reactant is left over?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s06_p15\" class=\"para\">Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less.<\/p>\n<p style=\"text-align: center\">[latex]5.00 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}}{85.4678 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb}} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}}{2 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}} \\times \\frac{24.3050 \\;\\text{g Mg}}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}} = 0.\\underline{711}53 \\;\\text{g Mg with 3 sig figs}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">[latex]3.44 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g MgCl}_2 \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2}{95.2104 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g MgCl}_2} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}}{1 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2} \\times \\frac{24.3050 \\;\\text{g Mg}}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Mg}} = 0.\\underline{878}15 \\;\\text{g Mg with 3 sig figs}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"ball-ch05_s06_p16\" class=\"para\">The 0.712 g of Mg is the lesser quantity, so the associated reactant\u20145.00 g of Rb\u2014is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl<sub class=\"subscript\">2<\/sub> reacted with the 5.00 g of Rb and then subtract the amount reacted from the original amount.<\/p>\n<p>[latex]5.00 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}}{85.4678 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Rb}} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2}{2 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Rb}} \\times \\frac{95.2104 \\;\\text{g MgCl}_2}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol MgCl}_2} = \\underline{2.78}498 \\;\\text{g magnesium chloride with 3 sig figs}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"ball-ch05_s06_p17\" class=\"para\">Because we started with 3.44 g of MgCl<sub class=\"subscript\">2<\/sub>, we have<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">3.44 g MgCl<sub class=\"subscript\">2<\/sub> \u2212 \u00a0<span style=\"text-decoration: underline\">2.78<\/span>498 g MgCl<sub class=\"subscript\">2<\/sub> reacted = 0.66 g MgCl<sub class=\"subscript\">2<\/sub> left<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s06_p18\" class=\"para\">Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent?<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/MGOS.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/MGOS-1.png\" alt=\"MGOS\" width=\"351\" height=\"68\" class=\"wp-image-3752 aligncenter\" \/><\/a><\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s06_p19\" class=\"para\">H<sub class=\"subscript\">2<\/sub>S is the limiting reagent; 1.5 g of MgO are left over.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"ball-ch05_s05_p06\" class=\"para\">A worker reacts 30.5 g of Zn with nitric acid and evapourates the remaining water to obtain 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. What are the theoretical yield, the actual yield, and the percent yield?<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 HNO<sub class=\"subscript\">3<\/sub>(aq) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s05_p07\" class=\"para\">A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.37 g\/mol) and Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> (189.38 g\/mol). In three steps, the mass-mass calculation is<\/p>\n<p>[latex]30.5 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Zn} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn}}{65.37 \\;\\rule[0.25ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g Zn}} \\times \\frac{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn(NO}_3 \\text{)}_2}{1 \\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn}} \\times \\frac{189.38 \\;\\text{g Zn(NO}_3 \\text{)}_2}{1 \\;\\rule[0.25ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol Zn(NO}_3 \\text{)}_2} = \\underline{88.3}6 \\;\\text{g zing nitrate with 3 sig figs}[\/latex]<\/p>\n<p id=\"ball-ch05_s05_p08\" class=\"para\">Thus, the theoretical yield is 88.4 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:<\/p>\n<p style=\"text-align: center\">[latex]\\text{percent yield} = (\\frac{65.2 \\;\\text{g Zn(NO}_3 \\text{)}_2}{\\underline{88.3}6 \\;\\text{g Zn(NO}_3 \\text{)}_2}) \\times 100\\%[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]= 73.8\\%[\/latex]<\/p>\n<p id=\"ball-ch05_s05_p09\" class=\"para\">The worker achieved almost three-fourths of the possible yield.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s05_p10\" class=\"para\">A synthesis produced 2.05 g of NH<sub class=\"subscript\">3<\/sub> from 16.5 g of N<sub class=\"subscript\">2<\/sub>. What is the theoretical yield and the percent yield?<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s05_p11\" class=\"para\">theoretical yield = 20.1 g; percent yield = 10.2%<\/p>\n<\/div>\n<div class=\"callout block\" id=\"ball-ch05_s05_n03\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification<\/h3>\n<p id=\"ball-ch05_s05_p12\" class=\"para\">Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.<\/p>\n<p id=\"ball-ch05_s05_p13\" class=\"para\">Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C<sub class=\"subscript\">13<\/sub>H<sub class=\"subscript\">21<\/sub>NO<sub class=\"subscript\">2<\/sub>; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:<\/p>\n<div class=\"informaltable\">\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td>impure albuterol \u2192\u00a0intermediate A<\/td>\n<td>percent yield = 70%<\/td>\n<\/tr>\n<tr>\n<td>intermediate A \u2192\u00a0intermediate B<\/td>\n<td>percent yield = 100%<\/td>\n<\/tr>\n<tr>\n<td>intermediate B \u2192\u00a0intermediate C<\/td>\n<td>percent yield = 40%<\/td>\n<\/tr>\n<tr>\n<td>intermediate C \u2192\u00a0intermediate D<\/td>\n<td>percent yield = 72%<\/td>\n<\/tr>\n<tr>\n<td>intermediate D \u2192\u00a0purified albuterol<\/td>\n<td>percent yield = 35%<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">overall percent yield = 70% \u00d7 100% \u00d7 40% \u00d7 72% \u00d7 35% = 7.5%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p id=\"ball-ch05_s05_p14\" class=\"para\">That is, only about <em class=\"emphasis\">one-fourteenth<\/em> of the original material was turned into the purified drug. This gives you one reason why some drugs are so expensive; a lot of material is lost in making a high-purity pharmaceutical.<\/p>\n<\/div>\n<h2 class=\"para\">Key Concepts and Summary<\/h2>\n<\/div>\n<p id=\"fs-idm5314032\">When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.<\/p>\n<\/section>\n<section id=\"fs-idp36036608\" class=\"key-equations\">\n<h2>Key Equations<\/h2>\n<ul id=\"fs-idp46607520\">\n<li>[latex]\\text{percent yield} = (\\frac{\\text{actual yield}}{\\text{theoretical yield}}) \\times 100\\%[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section id=\"fs-idm49138160\" class=\"exercises\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<p>1. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?<\/p>\n<p>2. A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was his percent yield?<\/p>\n<p>3. Freon-12, CCl<sub>2<\/sub>F<sub>2<\/sub>, is prepared from CCl<sub>4<\/sub> by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl<sub>2<\/sub>F<sub>2<\/sub> from 32.9 g of CCl<sub>4<\/sub>. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield.<\/p>\n<p>4. Toluene, C<sub>6<\/sub>H<sub>5<\/sub>CH<sub>3<\/sub>, is oxidized by air under carefully controlled conditions to benzoic acid, C<sub>6<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>H, which is used to prepare the food preservative sodium benzoate, C<sub>6<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?<\/p>\n<p id=\"fs-idp58383920\" style=\"text-align: center\">[latex]2\\text{C}_6\\text{H}_5\\text{CH}_3\\;+\\;3\\text{O}_2\\;{\\longrightarrow}\\;2\\text{C}_6\\text{H}_5\\text{CO}_2\\text{H}\\;+\\;2\\text{H}_2\\text{O}[\/latex]<\/p>\n<p>5. Outline the steps needed to solve the following problem, then do the calculations. Ether, (C<sub>2<\/sub>H<sub>5<\/sub>)<sub>2<\/sub>O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.<\/p>\n<p id=\"fs-idm120301824\" style=\"text-align: center\">[latex]2\\text{C}_2\\text{H}_5\\text{OH}\\;+\\;\\text{H}_2\\text{SO}_4\\;{\\longrightarrow}\\;(\\text{C}_2\\text{H}_5)_2\\;+\\;\\text{H}_2\\text{SO}_4{\\cdot}\\text{H}_2\\text{O}[\/latex]<\/p>\n<p id=\"fs-idp39307056\">What is the percent yield of ether if 1.17 L (d = 0.7134 g\/mL) is isolated from the reaction of 1.500 L of C<sub>2<\/sub>H<sub>5<\/sub>OH\u00a0(d = 0.7894 g\/mL)?<\/p>\n<p>6. Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H<sub>3<\/sub>PO<sub>4<\/sub> react according to the following chemical equation.<\/p>\n<p style=\"text-align: center\">[latex]2\\text{Cr}\\;+\\;2\\text{H}_3\\text{PO}_4\\;{\\longrightarrow}\\;2\\text{CrPO}_4\\;+\\;3\\text{H}_2[\/latex]<\/p>\n<p id=\"fs-idm522144\">Determine the limiting reactant.<\/p>\n<p>7. Uranium can be isolated from its ores by dissolving it as UO<sub>2<\/sub>(NO<sub>3<\/sub>)<sub>2<\/sub>, then separating it as solid UO<sub>2<\/sub>(C<sub>2<\/sub>O<sub>4<\/sub>)\u00b73H<sub>2<\/sub>O. Addition of 0.4031 g of sodium oxalate, Na<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub>, to a solution containing 1.481 g of uranyl nitrate, UO<sub>2<\/sub>(NO<sub>3<\/sub>)<sub>2<\/sub>, yields 1.073 g of solid UO<sub>2<\/sub>(C<sub>2<\/sub>O<sub>4<\/sub>)\u00b73H<sub>2<\/sub>O.<\/p>\n<p id=\"fs-idm72766640\" style=\"text-align: center\">[latex]\\text{Na}_2\\text{C}_2\\text{O}_4\\;+\\;\\text{UO}_2(\\text{NO}_3)_2\\;+\\;3\\text{H}_2\\text{O}\\;{\\longrightarrow}\\;\\text{UO}_2(\\text{C}_2\\text{O}_4){\\cdot}3\\text{H}_2\\text{O}\\;+\\;2\\text{NaNO}_3[\/latex]<\/p>\n<p id=\"fs-idp232896464\">Determine the limiting reactant and the percent yield of this reaction.<\/p>\n<p>8. How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_saccharin_img-2.jpg\" alt=\"A structural formula is shown. A hexagonal ring of 6 C atoms with alternating double bonds has single H atoms bonded to four consecutive C atoms on the left side of the ring. The two C atoms on the right side of the ring, which are joined by a double bond, are also included in a 5 atom ring to their right. The C atom of this pair that is nearest the top of the structure is singly bonded to a C atom at the top of the 5 atom ring which has an O atom double bonded above. An N atom is singly bonded to the lower right of this same C atom. The N atom has an H atom bonded to its right and to its lower left, it is bonded to an S atom. The S atom is connected to the second C atom that is shared in the two rings. The S atom is also double bonded to an O atom to its lower right and is double bonded to a second O atom directly below it.\" width=\"198\" height=\"155\" class=\"aligncenter\" \/><\/p>\n<p>9. Would you agree to buy 1 trillion (1,000,000,000,000) gold atoms for $5? Explain why or why not. Find the current price of gold at http:\/\/money.cnn.com\/data\/commodities\/ (1 troy ounce = 31.1 g)<\/p>\n<div class=\"qandaset block\" id=\"ball-ch05_s06_qs01\">\n<p>10. The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH<sub class=\"subscript\">3<\/sub>. What is the limiting reagent?<br \/>\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Limiting-Reagent.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Limiting-Reagent-1.png\" alt=\"Limiting Reagent\" width=\"276\" height=\"284\" class=\"wp-image-4659 aligncenter\" \/><\/a><\/p>\n<\/div>\n<div class=\"question\">\n<div class=\"informalfigure medium\"><span style=\"font-size: 1em\">11. Given the statement \u201c20.0 g of methane is burned in excess oxygen,\u201d is it obvious which reactant is the limiting reagent?<\/span><\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">12. Acetylene (C<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">) is formed by reacting 7.08 g of C and 4.92 g of H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">.<\/span><\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 C(s) +\u00a0H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s06_qs01_p10\" class=\"para\">What is the limiting reagent? How much of the other reactant is in excess?<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">13. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?<\/span><\/div>\n<div class=\"question\">\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-22-at-1.49.34-PM.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-22-at-1.49.34-PM-1.png\" alt=\"Screen Shot 2014-07-22 at 1.49.34 PM\" width=\"244\" height=\"49\" class=\"wp-image-3754 aligncenter\" \/><\/a><\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">14. To form the precipitate PbCl<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">, 2.88 g of NaCl and 7.21 g of Pb(NO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">)<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?<\/span><\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">15. What is the difference between the theoretical yield and the actual yield?<\/span><\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">16. A worker isolates 2.675 g of SiF<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\"> after reacting 2.339 g of SiO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> with HF. What are the theoretical yield and the actual yield?<\/span><\/div>\n<div class=\"question\">\n<div class=\"qandaset block\" id=\"ball-ch05_s05_qs01\">\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">SiO<sub class=\"subscript\">2<\/sub>(s) +\u00a04 HF(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0SiF<sub class=\"subscript\">4<\/sub>(g) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">17. A chemist decomposes 1.006 g of NaHCO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> and obtains 0.0334 g of Na<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">CO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">. What are the theoretical yield and the actual yield?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 NaHCO<sub class=\"subscript\">3<\/sub>(s) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0CO<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">18. What is the percent yield in Exercise 16?<\/span><\/p>\n<p><span style=\"font-size: 1em\">19. What is the percent yield in Exercise 17?<\/span><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><strong>Answers<\/strong><\/p>\n<p id=\"fs-idp22297728\">1. The limiting reactant is Cl<sub>2<\/sub>.<\/p>\n<p id=\"fs-idp46323152\">2. Percent yield = 31%<\/p>\n<p id=\"fs-idp59773152\">3. g CCl<sub>4<\/sub>\u00a0<span class=\"informalequation\"><span class=\"mathphrase\"><span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span><\/span><\/span> mol CCl<sub>4<\/sub>\u00a0<span class=\"informalequation\"><span class=\"mathphrase\"><span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span><\/span><\/span> mol CCl<sub>2<\/sub>F<sub>2<\/sub>\u00a0<span class=\"informalequation\"><span class=\"mathphrase\"><span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span><\/span><\/span> g CCl<sub>2<\/sub>F<sub>2<\/sub>, percent yield = 48.3%<\/p>\n<p id=\"fs-idm18984464\">4. percent yield = 91.3%<\/p>\n<p id=\"fs-idm48114608\">5. Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6%<\/p>\n<p id=\"fs-idm38791728\">6. The conversion needed is [latex]\\text{mol\\;Cr}\\;{\\longrightarrow}\\;\\text{mol\\;H}_3\\text{PO}_4[\/latex]. Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.<\/p>\n<p id=\"fs-idp74993536\">7. Na<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub> is the limiting reactant. percent yield = 86.6%<\/p>\n<p id=\"fs-idp81228400\">8. Only four molecules can be made.<\/p>\n<p id=\"fs-idp8960064\">9. This amount cannot be weighted by ordinary balances and is worthless.<\/p>\n<p>10. Nitrogen is the limiting reagent.<\/p>\n<p>11.\u00a0Yes; methane is the limiting reagent.<\/p>\n<p>12.\u00a0C is the limiting reagent; 4.33 g of H<sub class=\"subscript\">2<\/sub> are left over.<\/p>\n<p>13.\u00a0H<sub class=\"subscript\">2<\/sub>O is the limiting reagent; 25.9 g of P<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">6<\/sub> are left over.<\/p>\n<p>14.\u00a06.06 g of PbCl<sub class=\"subscript\">2<\/sub> are formed; 0.33 g of NaCl is left over.<\/p>\n<p>15. Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.<\/p>\n<p>16.\u00a0theoretical yield = 4.052 g; actual yield = 2.675 g<\/p>\n<p>17.\u00a0theoretical yield = 0.635 g; actual yield = 0.0334 g<\/p>\n<p>18.\u00a066.02%<\/p>\n<p>19.\u00a05.26%<\/p>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<p><strong>actual yield:\u00a0<\/strong>amount of product formed in a reaction<\/p>\n<p><strong>excess reactant:\u00a0<\/strong>reactant present in an amount greater than required by the reaction stoichiometry<\/p>\n<p><strong>limiting reactant:\u00a0<\/strong>reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated<\/p>\n<p><strong>percent yield:\u00a0<\/strong>measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield<\/p>\n<p><strong>theoretical yield:\u00a0<\/strong>amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry<\/p>\n<\/div>\n","protected":false},"author":330,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"7.2 Limiting Reagent and Reaction Yields","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[54],"class_list":["post-1435","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1405,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1435","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/users\/330"}],"version-history":[{"count":19,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1435\/revisions"}],"predecessor-version":[{"id":4757,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1435\/revisions\/4757"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/1405"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1435\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=1435"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=1435"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=1435"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=1435"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}